# Parabola - continued

## Construct a parabola, given...

That is in fact construct the focus and the directrix.

### Given axis and two points

This is possible only if AB is neither parallel (no solution) no perpendicular (infinitely many) to axis.
We immediately get two other points C and D symetric of A and B from axis.
Considering a 5th point as point at infinity in direction of axis, the Pascal theorem gives a construction of the tangent in A :
AB and CD intersect in O.
AC and parallel from B to axis (line B to P∞) intersect in U.
OU and parallel from D to axis (DP∞) intersect in V.
AV is the tangent in A to parabola.
Angle reflexion of the parallel to axis on this tangent intersects axis in the focus F.
The directrix is the perpendicular to axis from point H, symetric of F from the tangent.

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### 3 points and direction of axis

It is really a conic section defined by 5 points, the 3 given points and two overlapping points at infinity in direction of axis (parabola is tangent to the line at infinity).
We then construct immediately a 4th point D at a finite distance as the line through the midpoints I and J of two parallel secants AB and CD, is parallel to axis.
Considering then the 5th point as the infinity point P∞ in direction of axis, Pascal theorem gives a construction of the tangents in A and B :
BC and AD intersect in U.
AB and parallel to axis from D (DP∞) intersect in V
UV and parallel to axis from C (CP∞) intersect in W
AW is the tangent in A.
Do the same for constructing the tangent in B, reversing the role of A and B.
Reflexions on these tangents of rays parallel to axis intersect in focus F.
Which gives the axis itself, then the directrix through H symetric of F from the tangent in A, or just AH=AF.

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### 4 points

There are generally two solutions. If ABCD is not convex there are no solutions.
A construction by Newton :
The diagonals AC and BD intersect in O.
Circle with diameter AC intersects the perpendicular from O to AC in point K. OM = OM' = OK is copied to AC line. OM² = OM'² = OA.OC
In the same way, circle with diameter BD intersects the perpendicular from O to BD in K' and ON = OK' is copied on BD line. ON² = OB.OD
MN and M'N are the directions of axis of the two parabolas going through points ABCD.

The Pascal theorem and the point at infinity in direction MN as a 5th point, give a construction of the tangents to parabola in A and D.
Parallel to MN from C intersects AD in U. UO intersects the parallel from B to MN in point V. AV is the tangent in A.
The tangent in D is similarily constructed, exchanging A↔D and B↔C, DV' is the tangent in D.

The parallels to MN from A and D reflect on the tangents into rays going at focus F.
The symetric H of F from the tangent in A (or just AH = AF) is on the directrix, which is then the perpendicular to MN from H.

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### 3 tangents and direction of axis

The perpendicular projection of focus on the 3 tangents are on the tangent at apex, hence are in line.
The tangent at apex is then the Simson line of the focus for triangle ABC, hence the focus is on the circumcircle of ABC.
Hence the construction :

from C draw the perpendicular to axis direction. It intersects the circumcircle in P.
The perpendicular from P to AB intersects the circumcircle at the focus F.
The symetric of F from AB gives a point K on the directrix, the perpendicular from K to axis direction is then the directrix.
The parallel from K to axis direction gives the contact point T on AB.
Symetric points of F from BC and AC give the other contact points.

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The points on the directrix define the directrix as the Steiner line of focus. This line (that is the directrix) goes at orthocenter H of ABC (as Steiner line).
This gives another construction : the directrix is the perpendicular to axis direction from the orthocenter H of ABC.
Focus is then the intersection point of the symetric lines of directrix from the sides (the tangents).

### 4 tangents

As previously, the focus is on the circumcircle of ABC, and also on circumcircles of ADE, CDG and BEG. These 4 circles have a common point, the Miquel point of complete quadrilateral ABCDEG.
Hence the construction : circumcircles to ABC and ADE intersect in F.
Symetric points H and K of F from the two tangents AB and AD define the directrix.
Perpendiculars to the directrix from these points give the contact points T1 and T4, and similarily for the two other tangents.

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The orthocenters of the 4 triangles are in line on the Steiner line.
This gives a direct construction of the directrix by constructing the orthocenters H1 and H2 of triangles CDG and EBG,
then the symetric lines of directrix from the sides AB and DE intersect at focus F.

### And then...

There are other interesting cases with a mix of known points and/or tangents. Not all are constructible with compass and straightedge.
For instance, given the apex and two points, this is enough data and the parabola can be determined.
Unfortunately, this requires a 3rd degree equation, hence generally not constructible with compass and straightedge.
More details about this case.

And also : construct a parabola with given focus.